Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resistance. Find

Question

Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resistance. Find the current ????1 through R1 and the potential difference V2 across R2 .

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Thu Thảo 3 years 2021-08-17T09:51:01+00:00 1 Answers 12 views 0

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  1. Delwyn
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    2021-08-17T09:52:50+00:00
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    Answer:

    (a) 2.33 A

    (b) 15.075 V

    Explanation:

    From the question,

    The total resistance (Rt) = R1+R2 = 3.85+6.47

    R(t) = 10.32 ohms.

    Applying ohm’s law,

    V = IR(t)……….equation 1

    Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.

    Note: Since both resistors are connected in series, the current flowing through them is the same.

    Therefore,

    I = V/R(t)…………. Equation 2

    Given: V = 24 V, R(t) = 10.32 ohms

    Substitute these values into equation 2

    I = 24/10.32

    I = 2.33 A.

    Hence the current through R1 = 2.33 A.

    V2 = IR2………….. Equation 3

    V2 = 2.33(6.47)

    V2 = 15.075 V

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